Neco Mathematics Essay Answer 2018
NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers – related to these are NECO past questions 2016,2016/2017 NECO mathematics answers,2015 NCO past questions and answers, NECO mathematics past questions and answers, NECO past questions 2016 mathematics and NECO mathematics question and answer 2015
In this post are researched regularly repeated NECO Mathematics Questions – Objectives and Theory questions and answers for 2018 June/July SSCE exam. Get them in your finger tips and you are sure of a high grade in this exams. NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers.
NECO 2018 Mathematics Questions & Answers
WHAT NECO STANDS FOR
The National Examinations Council – NECO is a Nigerian Government owned examination body that conducts the Senior Secondary Certificate Examination and the General Certificate in Education in June/July and December/January respectively. These are entry qualification examinations for higher education institutions in Nigeria.
NECO was created in April 1999 with a mandated to take over the responsibilities of the National Board of Education Measurement (NBEM).
NECO therefore conducts Senior Secondary Certificate Examination (internal and external), Junior Secondary Certificate Examination (JSCE) for Federal Unity College and other Federal secondary schools and Private secondary schools, and National Common Entrance Examination for primary 6 pupils interested in admission into Federal Unity Colleges.
The Senior Secondary Certificate Examinations (internal and external) are entry qualification examinations for higher education institutions in Nigeria.
NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives
NECO RESULTS & HIGHER SCHOOL ADMISSION
As you know, no University, Polytechnic, or college of education in Nigeria would admit you into any of their courses without a good grade in any of these examinations. That is why over three million candidates register and take these subject exams annually.
Consequently, in NECO 2018 Mathematics Questions & Answers/2018 NECO Mathematics – Theory And Objectives – Questions And Answers, our goal is to help you achieve this your life dream. Listed here are Mathmatics, Objective and Theory questions and answers relevant to 2018 June/July Mathematics Subject examination.
2018 NECO Mathematics – Theory And Objectives
VERIFIED NECO 2018 MATHEMATICS QUESTIONS AND ANSWERS
Paper III: Objective – General Mathematics
Paper II: Essay – General Mathematics
1-10 deacdacbbe
11-20 cbaeebdacc
21-30 deedacbbce
31-40 abdcebaeec
41-50 caadedbaea
51-60 edeacbacdd
1a)
Tabulate
x- 1,2,3,4
1- 1,2,3,4
2- 2_, 4, 0_ ,2_
3- 3, 0_, 3, 0_
4- 4_, 2_, 0_, 4
1b)
I = PRT/100,
p=N15000
R=10% and I=3years
A = P I where I = 15000*10*3/100=N4500
A=4500 15000 =N19500
2a)
using sine rule
b/sin20 = 6/sin30
bsin30 = 6sin120
b 6sin120/sin30
b = 6×0.2511/0.4540
b = 5.7063/0.4540
b = 12.57 ≠ 12.6cm
2bi)
the diagram is equivalent triangles.
where
|AX|/|BC| = |BY|/|AC| = |XY|/|YC|
XY = 9, BY = 7
YC = 18-7=11
9/11 = 7/|AC|
9|AC| = 77
|AC| = 77/9
|AC| = 8cm
2bii)
XY/AB = BY/AC
9/|AB| = 7/8.6
|AB| = 9×8.6/7
|AB| = 11cm
Read more;
3)
let the son age be x
man=5x
son=x
4yrs ago; the man age = 5x – 4
the son age = x – 4
the product of their ages
(5x – 4)(x – 4) =448
4a)
volume of fuel = cross-sectional area of X depth of fuel rectangular tank
30,000litres = 7.5*4.2*d m^3
but; 1000litres =1m^3
therefore;30(M^3) = 7.5*4.2*d(M^3)
30=31.5d
====> d = 30/31.5 = 0.95(2d.p)
4b)
to fill the tank/volume of fuel needed = 7.5*4.2*1.2 = 37.8m^3 = 37,800
litres addition fuel = 37,800-30,000 = 7,800
litres therefore, 7,800
more litres would be needed
================
5a)
sector for building project =48000/144000*360 =120degree
sector for education = 32,000/144000*360=80degree
sector for saving = 19200/144000*360=48degree
sector for maintenance = 12000/144000*360=
30degree
sector for miscellaneous = 7200/144000*360=18degree
sector for food items = 360- (120 80 48 30 18) =360-296 =64degree
5b)
amount spent=144000- [48,000 32000 19200 12000 7200] =144000-118400 =N25600
===============
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7a)
3²ⁿ ¹ — 4(3ⁿ ¹) 9 = 0
3²ⁿ × 3 — 4(3ⁿ× 3¹) 9 = 0
(3ⁿ)² × 3 — 4(30ⁿ× ) 9 = 0
Let 3ⁿ= x
3x² — 4 × 3 × x 9 = 0
3x² — 12x 9 = 0
Divide all by 3
3x²/3 — 12x/3 9/3 = 0
x² — 4x 3 = 0
x² — 3x — x 3 = 0
x(x—3) -1(x—3) = 0
(x—3)(x—3) = 0
x—3 = 0 or x—1 = 0
x = 3, x = 1
Substitute x = 3
3ⁿ = 3 or 3ⁿ = 1
3ⁿ = 3¹ or 3ⁿ= 3°
n = 1 or n= 0
7b)
log(x^2 4) = 2 logx – log^20
log(x^2 4) = log^100 = log^x – log^20
(x^2 4) = log(xx)
x^2 4 = 5x
x^2-5x 4 = 0
x^2-4x – x 4 = 0
x(x-4) – 1(x-4) = 0
(x-1)(x-4) = 0
x-1 = 0 or x-4 = 0
x = 1 or 4
8)
|BC|² = |BD| + |CD|²
13² = BD² + 5²
169 = BD² + 25
√BD² = √144
BD= 12m
Both OB = OD and OB + OD
= 2OB/2 = 12cm/2
OB = 6cm
8b) Circumference = 2πr
r = 6cm, π = 22/7
= 22 × 2 × 6/7cm
= 264/7cm
= 37.7cm to 1 decimal places
=================
9a) Let the digits be y
10(5 + y)+y =3y(5 +y) —14
50 + 10y + y = 3y(5+y) —14
50 + 11y = 15y + 3y² — 14
3y² + 15y — 11y — 50 — 14 = 0
3y² + 4y — 64 = 0
(3y² — 12y) + (16y — 64) = 0
3y(y — 4)(3y + 16) = 0
y —4 = 0 or 3y + 16 = 0
y = 4 or —16/3
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9b)
3—2x/4 + 2x—3″3
= 3(3—2x) + 4(2x—3)/12
= 9—6x + 8x—12/12
= 2x—3/12
10a)
y=(2x^2 + 3)^5
let U=2x^2 + 3
Y=u^5
du/dx = 4x
dy/du = 5u^4
dy/du = (2x^2 + 3)^4
dy/dx = du/dx dy/du
dy/dx = 4x.5(2x^2 + 3)^4
dy/dx = 20x(2x^2 + 3)^4
10b)
y=3x^2 + 2x +5
dy/dx =6x + 2
dy/dx =6(3) +2
dy/dx =18+2
dy/dx =20
10c)
R-W=Wv^2/gx
Wv^2=gx(R-W)
Wv^2=gRx-Wgx
Wv^2+Wgx=gRx
W(v^2 + gx) =gRx
W=gRx/V^2 + gx
R=2, g=10, x=3/2, V=3
W= 10*2*3/2/3^2 + 10*3/5
W=30/9+15
W=30/24
W=5/4
Good luck.
2018 NECO Mathematics – Theory And Objectives
RELEVANT PAST QUESTIONS
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Neco Mathematics Essay Answer 2018
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